Pembahasan Soal Ujian Nasional Limit Fungsi Aljabar

 Pembahasan soal-soal Ujian Nasional (UN) SMA bidang studi Matematika IPA untuk pokok bahasan Limit Fungsi Aljabar.

1. EBT 2003
Nilai dari ${}_{x\to 2}^{lim}\frac{4-x^2}{3-\sqrt{x^2+5}}$ = ...
A.  −12
B.  −6
C.  0
D.  6
E.  12

Pembahasan :
$\begin{array}{cc}\underset{x\to 2}{lim}\frac{4-x^2}{3-\sqrt{x^2+5}}& =\underset{x\to 2}{lim}\frac{4-x^2}{3-\sqrt{x^2+5}}\cdot \frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}\\ & =\underset{x\to 2}{lim}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{9-\left(x^2+5\right)}\\ & =\underset{x\to 2}{lim}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{4-x^2}\\ & =\underset{x\to 2}{lim}\left(3+\sqrt{x^2+5}\right)\\ & =3+\sqrt{2^2+5}\\ & =6\end{array}$

Jawaban : D



2. UN 2004
Nilai ${}_{x\to 2}^{lim}\left(\frac{2}{x^2-4}-\frac{3}{x^2+2x-8}\right)$ = ...
A.  $-\frac{7}{12}$
B.  $-\frac{1}{4}$
C.  $-\frac{1}{12}$
D.  $-\frac{1}{24}$
E.  0

Pembahasan :
$\begin{array}{cc}& \underset{x\to 2}{lim}\left(\frac{2}{x^2-4}-\frac{3}{x^2+2x-8}\right)\\ & =\underset{x\to 2}{lim}\left(\frac{2}{(x+2)(x-2)}-\frac{3}{(x+4)(x-2)}\right)\\ & =\underset{x\to 2}{lim}\frac{2(x+4)-3(x+2)}{(x+2)(x-2)(x+4)}\\ & =\underset{x\to 2}{lim}\frac{-(x-2)}{(x+2)(x-2)(x+4)}\\ & =\underset{x\to 2}{lim}\frac{-1}{(x+2)(x+4)}\\ & =\frac{-1}{(2+2)(2+4)}\\ & =-\frac{1}{24}\end{array}$

Jawaban : D



3. UN 2006
Nilai ${}_{x\to 6}^{lim}\frac{\sqrt{3x-2}-\sqrt{2x+4}}{x-6}$ = ...
A.  $-\frac{1}{4}$
B.  $-\frac{1}{8}$
C.  0
D.  $\frac{1}{8}$
E.  $\frac{1}{4}$

Pembahasan :
$\begin{array}{cc}& \underset{x\to 6}{lim}\frac{\sqrt{3x-2}-\sqrt{2x+4}}{x-6}\\ & =\underset{x\to 6}{lim}\frac{\sqrt{3x-2}-\sqrt{2x+4}}{x-6}\cdot \frac{\sqrt{3x-2}+\sqrt{2x+4}}{\sqrt{3x-2}+\sqrt{2x+4}}\\ & =\underset{x\to 6}{lim}\frac{(3x-2)-(2x+4)}{(x-6)\left(\sqrt{3x-2}+\sqrt{2x+4}\right)}\\ & =\underset{x\to 6}{lim}\frac{(x-6)}{(x-6)\left(\sqrt{3x-2}+\sqrt{2x+4}\right)}\\ & =\underset{x\to 6}{lim}\frac{1}{\sqrt{3x-2}+\sqrt{2x+4}}\\ & =\frac{1}{\sqrt{3\left(6\right)-2}+\sqrt{2\left(6\right)+4}}\\ & =\frac{1}{8}\end{array}$

Jawaban : D



4. UN 2007
Nilai ${}_{x\to 3}^{lim}\frac{x^2-x-6}{4-\sqrt{5x+1}}$ = ...
A.  −8
B.  −6
C.  6
D.  8
E.  ∞

Pembahasan :
$\begin{array}{cc}\underset{x\to 3}{lim}\frac{x^2-x-6}{4-\sqrt{5x+1}}& =\underset{x\to 3}{lim}\frac{x^2-x-6}{4-\sqrt{5x+1}}\cdot \frac{4+\sqrt{5x+1}}{4+\sqrt{5x+1}}\\ & =\underset{x\to 3}{lim}\frac{(x^2-x-6)\left(4+\sqrt{5x+1}\right)}{16-(5x+1)}\\ & =\underset{x\to 3}{lim}\frac{(x-3)(x+2)\left(4+\sqrt{5x+1}\right)}{-5(x-3)}\\ & =\underset{x\to 3}{lim}\frac{(x+2)\left(4+\sqrt{5x+1}\right)}{-5}\\ & =\frac{(3+2)\left(4+\sqrt{5\left(3\right)+1}\right)}{-5}\\ & =-8\end{array}$

Jawaban : A



5. UN 2008
Nilai ${}_{x\to 2}^{lim}\frac{x^3-4x}{x-2}$ = ...
A.  32
B.  16
C.  8
D.  4
E.  2

Pembahasan :
$\begin{array}{cc}\underset{x\to 2}{lim}\frac{x^3-4x}{x-2}& =\underset{x\to 2}{lim}\frac{x\left(x^2-4\right)}{x-2}\\ & =\underset{x\to 2}{lim}\frac{x(x+2)(x-2)}{x-2}\\ & =\underset{x\to 2}{lim}x(x+2)\\ & =2(2+2)\\ & =8\end{array}$

Jawaban : C



6. UN 2009
Nilai ${}_{x\to 3}^{lim}\frac{x^2-9}{\sqrt{10+2x}-(x+1)}$ = ...
A.  −8
B.  −6
C.  4
D.  6
E.  8

Pembahasan :
$\begin{array}{cc}& \underset{x\to 3}{lim}\frac{x^2-9}{\sqrt{10+2x}-(x+1)}\\ & =\underset{x\to 3}{lim}\frac{x^2-9}{\sqrt{10+2x}-(x+1)}\cdot \frac{\sqrt{10+2x}+(x+1)}{\sqrt{10+2x}+(x+1)}\\ & =\underset{x\to 3}{lim}\frac{\left(x^2-9\right)\left(\sqrt{10+2x}+(x+1)\right)}{10+2x-(x^2+2x+1)}\\ & =\underset{x\to 3}{lim}\frac{\left(x^2-9\right)\left(\sqrt{10+2x}+(x+1)\right)}{-\left(x^2-9\right)}\\ & =\underset{x\to 3}{lim}\frac{\sqrt{10+2x}+(x+1)}{-1}\\ & =\frac{\sqrt{10+2\left(3\right)}+(3+1)}{-1}\\ & =-8\end{array}$

Jawaban : A



7. UN 2010
Nilai ${}_{x\to 0}^{lim}\left(\frac{4x}{\sqrt{1-2x}-\sqrt{1+2x}}\right)$ = ...
A.  −2
B.  0
C.  1
D.  2
E.  4

Pembahasan :
$\begin{array}{cc}& \underset{x\to 0}{lim}\frac{4x}{\sqrt{1-2x}-\sqrt{1+2x}}\\ & =\underset{x\to 0}{lim}\frac{4x}{\sqrt{1-2x}-\sqrt{1+2x}}\cdot \frac{\sqrt{1-2x}+\sqrt{1+2x}}{\sqrt{1-2x}+\sqrt{1+2x}}\\ & =\underset{x\to 0}{lim}\frac{4x\left(\sqrt{1-2x}+\sqrt{1+2x}\right)}{(1-2x)-(1+2x)}\\ & =\underset{x\to 0}{lim}\frac{4x\left(\sqrt{1-2x}+\sqrt{1+2x}\right)}{-4x}\\ & =\underset{x\to 0}{lim}\frac{\sqrt{1-2x}+\sqrt{1+2x}}{-1}\\ & =\frac{\sqrt{1-2\left(0\right)}+\sqrt{1+2\left(0\right)}}{-1}\\ & =-2\end{array}$

Jawaban : A



8. UN 2011
Nilai ${}_{x\to 4}^{lim}\frac{(x-4)}{\sqrt{x}-2}$ = ...
A.  0
B.  4
C.  8
D.  12
E.  16

Pembahasan :
$\begin{array}{cc}\underset{x\to 4}{lim}\frac{x-4}{\sqrt{x}-2}& =\underset{x\to 4}{lim}\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\\ & =\underset{x\to 4}{lim}\left(\sqrt{x}+2\right)\\ & =\sqrt{4}+2\\ & =4\end{array}$

Jawaban : B



9. UN 2012
Nilai ${}_{x\to 3}^{lim}\frac{2-\sqrt{x+1}}{x-3}=...$
A.  $-\frac{1}{4}$
B.  $-\frac{1}{2}$
C.  1
D.  2
E.  4

Pembahasan :
$\begin{array}{cc}\underset{x\to 3}{lim}\frac{2-\sqrt{x+1}}{x-3}& =\underset{x\to 3}{lim}\frac{2-\sqrt{x+1}}{x-3}\cdot \frac{2+\sqrt{x+1}}{2+\sqrt{x+1}}\\ & =\underset{x\to 3}{lim}\frac{4-(x+1)}{(x-3)\left(2+\sqrt{x+1}\right)}\\ & =\underset{x\to 3}{lim}\frac{-(x-3)}{(x-3)\left(2+\sqrt{x+1}\right)}\\ & =\underset{x\to 3}{lim}\frac{-1}{2+\sqrt{x+1}}\\ & =\frac{-1}{2+\sqrt{3+1}}\\ & =-\frac{1}{4}\end{array}$

Jawaban : A



10. UN 2013
Nilai dari $\begin{array}{c}\underset{x\to \infty }{lim}\left((2x-1)-\sqrt{4x^2-6x-5}\right)=...\end{array}$
A.  4
B.  2
C.  1
D.  $\frac{1}{2}$
E.  $\frac{1}{4}$

Pembahasan :
Misalkan  $\begin{array}{c}\underset{x\to \infty }{lim}\left((2x-1)-\sqrt{4x^2-6x-5}\right)=L\end{array}$

$\begin{array}{cc}L& =\underset{x\to \infty }{lim}\left(\sqrt{(2x-1{)}^2}-\sqrt{4x^2-6x-5}\right)\\ & =\underset{x\to \infty }{lim}\left(\sqrt{4x^2-4x+1}-\sqrt{4x^2-6x-5}\right)\end{array}$

a = 4,  b = -4,  c = 1
p = 4,  q = -6,  r = -5

Karena a = p maka berlaku
$\begin{array}{c}L=\frac{b-q}{2\sqrt{a}}=\frac{-4-(-6)}{2\sqrt{4}}=\frac{2}{4}=\frac{1}{2}\end{array}$

Jawaban : D



11. UN 2013
Nilai dari $\begin{array}{c}\underset{x\to \infty }{lim}\frac{\sqrt{5-4x+3x^2}+\sqrt{4-3x+3x^2}}{2x}=...\end{array}$
A.  0
B.  $\frac{1}{3}$√3
C.  √3
D.  2√3
E.  ∞

Pembahasan :
Karena pangkat tertinggi pembilang sama dengan pangkat tertinggi penyebut, maka nilai limit tak hingga diatas sama dengan koefisien pangkat tertinggi pembilang dibagi koefisien pangkat tertinggi penyebut.

$\begin{array}{cc}\underset{x\to \infty }{lim}\frac{\sqrt{5-4x+3x^2}+\sqrt{4-3x+3x^2}}{2x}& =\frac{\sqrt{3}+\sqrt{3}}{2}\\ & =\sqrt{3}\end{array}$

Jawaban : C



12. UN 2014
Nilai $\begin{array}{c}\underset{x\to \infty }{lim}\left(\sqrt{9x^2+6x-2}-3x+1\right)=...\end{array}$
A.  5
B.  4
C.  3
D.  2
E.  1

Pembahasan :
Misalkan  $\begin{array}{c}\underset{x\to \infty }{lim}\left(\sqrt{9x^2+6x-2}-3x+1\right)=L\end{array}$

$\begin{array}{cc}L& =\underset{x\to \infty }{lim}\left(\sqrt{9x^2+6x-2}-(3x-1)\right)\\ & =\underset{x\to \infty }{lim}\left(\sqrt{9x^2+6x-2}-\sqrt{(3x-1{)}^2}\right)\\ & =\underset{x\to \infty }{lim}\left(\sqrt{9x^2+6x-2}-\sqrt{9x^2-6x+1}\right)\end{array}$

a = 9,  b = 6,  c = -2
p = 9,  q = -6,  r = 1

Karena a = p, maka berlaku :
$\begin{array}{c}L=\frac{b-q}{2\sqrt{a}}=\frac{6-(-6)}{2\sqrt{9}}=\frac{12}{6}=2\end{array}$

Jawaban : D



13. UN 2016
Nilai dari $\begin{array}{c}\underset{x\to \infty }{lim}\left(\sqrt{4x^2+4x-3}-(2x-5)\right)=...\end{array}$
A.  −6
B.  −4
C.  −1
D.  4
E.  6

Pembahasan :
Misalkan  $\begin{array}{c}\underset{x\to \infty }{lim}\left(\sqrt{4x^2+4x-3}-(2x-5)\right)=L\end{array}$

$\begin{array}{cc}L& =\underset{x\to \infty }{lim}\left(\sqrt{4x^2+4x-3}-\sqrt{(2x-5{)}^2}\right)\\ & =\underset{x\to \infty }{lim}\left(\sqrt{4x^2+4x-3}-\sqrt{4x^2-20x+25}\right)\end{array}$

a = 4,  b = 4,  c = -3
p = 4,  q = -20,  r = 25

Karena a = p maka berlaku
$\begin{array}{c}L=\frac{b-q}{2\sqrt{a}}=\frac{4-(-20)}{2\sqrt{4}}=\frac{24}{4}=6\end{array}$

Jawaban : E

sumber: https://smatika.blogspot.com/

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